Answer to Question #267817 in Discrete Mathematics for fevw

Question

2.     For each function below (from Z to Z), indicate whether the function is onto, one-to-one, neither or both. If the function is not onto or not one-to-one, give an example showing why. If the function is bijective, find and show its inverse.

a)  f (x) = 3x – 1

b)  f(x) = x² + 2x­­ + 1

c)  f(x,y)= 2y -3x

Answer 1

function f: X -> Y is called:

• onto $\iff$ $\exists y\in Y \exists x\in X:y=f(x)$
• one-to-one $\iff (\forall a,b\in X: f(a)=f(b)\iff a=b)$
• bijection $\iff$ each element of X is paired with exactly one element of Y, and each element of Y is paired with exactly one element of X. $bijection\iff onto\land one-to-one$

a)  f (x) = 3x – 1

Since this function is defined for every $x\in Z$ continious and monotonically increasing, than for any X it has different f(x), which means it is one-to-one. Lets put $f(x) = 1\implies 1 = 3x - 1 \implies x={\frac 2 3} \notin Z$ , which means it is not onto.

This function is one-to-one and not onto

b) f(x) = x² + 2x­­ + 1

Since for f(x) = 1 both x = 0 and x = -2 are fit, than f(x) is not one-to-one. Also for f(x) = -1 we have

x² + 2x­­ + 2 = 0 $\implies D = b^2-4ac=4-8 = -4<0$ , which means there is no such $x\isin Z$

So, this function is neither onto nor onto.

c) f(x,y)= 2y -3x

Since for f(x,y) = -1 both x = 1, y = 1 and x = -1, y = -2 are fit, then f(x,y) is not one-to-one.

Let $a\in Z: a=2y-3x\implies y={\frac {3x+a} 2}$ . Then we can put x = a and receive y = 2a. $a\in Z\implies x=a \isin Z, y=2a\in Z$ , which means f(x, y) is onto

This function is onto and not one-to-one